[精品]列主元高斯消往法、lu三角分化法、龙贝格(romberg)算法、最小二乘法的matlab法度模范及运转结果
1(
列主元高斯消去法
M文件
function[x]=gauss(a,b)
n=length(a);
x=zeros(n,1);
a=[a b];
for k=1:n-1
max=k;
for i=k+1:n
if a(i,k)>a(max,k)
max=i;
end
end
temp=a(k,k:n+1);
a(k,k:n+1)=a(max,k:n+1);
a(max,k:n+1)=temp;
for i=k+1:n
a(i,k)=-a(i,k)/a(k,k);
a(i,k+1:n+1)=a(i,k+1:n+1)+a(i,k)*a(k,k+1:n+1);
end
end
x(n,1)=a(n,n+1)/a(n,n);
for i=n-1:-1:1
sum=0;
for j=i+1:n
sum=sum+x(j,1)*a(i,j);
end
x(i,1)=(a(i,n+1)-sum)/a(i,i);
end
Matlab运行结果
2(LU三角分解法
M文件
function y=LU(A,B); n=length(A); A=[A B]; for k=1:n-1;
for i=k:n;
if(abs(A(i,k))==max(abs(A(k:n,k))))
P(k)=i;
temp=A(k,:);
A(k,:)=A(i,:);
A(i,:)=temp;
end
end
for j=k+1:n;
A(j,k)=A(j,k)/A(k,k);
A(j,k+1:n+1)=A(j,k+1:n+1)-A(j,k)*A(k,k+1:n+1);
end
end
P(n)=n;
L(1,1)=1;
L(2:n,1)=A(2:n,1);
L(1,2:n)=0;
U(1,1)=A(1,1);
U(2:n,1)=0;
U(1,2:n)=A(1,2:n);
for i=2:n;
L(i,1:i-1)=A(i,1:i-1);
L(i,i)=1;
L(i,i+1:n)=0;
U(i,1:i-1)=0;
U(i,i:n)=A(i,i:n); end
x(n) = A(n,n+1)/U(n,n); for k = n-1:-1:1
x(k)=A(k,n+1);
for p=n:-1:k+1;
x(k) = x(k)-U(k,p)*x(p); end
x(k)=x(k)/U(k,k); end
x
L
U
P
End
Matlab运行结果
3(龙贝格(Romberg)算法
M文件
function[t]=romberg(f,a,b,e)
t=zeros(15,4);
t(1,1)=(b-a)/2*(f(a)+f(b));
for k=2:4
sum=0;
for i=1:2^(k-2)
sum=sum+f(a+(2*i-1)*(b-a)/2^(k-1));
end
t(k,1)=0.5*t(k-1,1)+(b-a)/2^(k-1)*sum;
for i=2:k
t(k,i)=(4^(i-1)*t(k,i-1)-t(k-1,i-1))/(4^(i-1)-1);
end
end
for k=5:15
sum=0;
for i=1:2^(k-2)
sum=sum+f(a+(2*i-1)*(b-a)/2^(k-1));
end
t(k,1)=0.5*t(k-1,1)+(b-a)/2^(k-1)*sum;
for i=2:4
t(k,i)=(4^(i-1)*t(k,i-1)-t(k-1,i-1))/(4^(i-1)-1);
end
if k>6
if abs(t(k,4)-t(k-1,4))
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
',num2str(t(k,4))]);
break;
end
end
end
if k>=15
disp(['溢出']);
end
Matlab运行结果
4(最小二乘法
M文件
function[a,max,det]=zuixiaoerchengfa(x,y,r)
n=length(x);
c=ones(n,r+1);
for i=2:r+1
for j=1:n
c(j,i)=x(j)^(i-1);
end
end
A=c'*c;
b=c'*y';
a=inv(A)*b;
det=0;
max=0;
for i=1:n
sum=a(1);
for j=2:r+1
sum=sum+a(j)*x(i)^(j-1);
end
cc=abs(y(i)-sum);
if cc>max
max=cc;
end
det=det+cc^2;
end
det=sqrt(det);
Matlab运行结果
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